Pat05 ppt 0105

1. Copyright © 2016, 2012 Pearson Education, Inc. 1 -1
Chapter 1
Graphs,
Functions,
and Models

2. Copyright © 2016, 2012 Pearson Education, Inc. 1 -2
Section
1.1 Introduction to Graphing
1.2 Functions and Graphs
1.3 Linear Functions, Slope, and Applications
1.4 Equations of Lines and Modeling
1.5 Linear Equations, Functions, Zeros and
Applications
1.6 Solving Linear Inequalities

3. Copyright © 2016, 2012 Pearson Education, Inc. 1 -3
1.5
Linear Equations, Functions, Zeros,
and Applications
 Solve linear equations.
 Solve applied problems using linear models.
 Find zeros of linear functions.

4. Copyright © 2016, 2012 Pearson Education, Inc. 1 -4
Equations and Solutions
An equation is a statement that two expressions are
equal.
To solve an equation in one variable is to find all the
values of the variable that make the equation true.
Each of these values is a solution of the equation.
The set of all solutions of an equation is its solution
set.
Some examples of equations in one variable are
2x  3  5, 3 x 1  4x  5,
x  3
x  4
 1,
and x2
 3x  2  0.

5. Copyright © 2016, 2012 Pearson Education, Inc. 1 -5
Linear Equations
A linear equation in one variable is an equation
that can be expressed in the form mx + b = 0, where
m and b are real numbers and m ≠ 0.

6. Copyright © 2016, 2012 Pearson Education, Inc. 1 -6
Equivalent Equations
Equations that have the same solution set are
equivalent equations.
For example, 2x + 3 = 5 and x = 1 are equivalent
equations because 1 is the solution of each equation.
On the other hand, x2 – 3x + 2 = 0 and x = 1 are not
equivalent equations because 1 and 2 are both
solutions of x2 – 3x + 2 = 0 but 2 is not a solution of
x = 1.

7. Copyright © 2016, 2012 Pearson Education, Inc. 1 -7
Equation-Solving Principles
For any real numbers a, b, and c:
The Addition Principle:
If a = b is true, then a + c = b + c is true.
The Multiplication Principle:
If a = b is true, then ac = bc is true.

8. Copyright © 2016, 2012 Pearson Education, Inc. 1 -8
Example
Solve:
Solution: Start by multiplying both sides of the
equation by the LCD to clear the equation of fractions.
3
4
x 1 
7
5
20
3
4
x 1



  20
7
5
20
3
4
x  20 1  28
15x  20  28
15x  20  20  28  20
15x  48
15x
15

48
15
x 
48
15
x 
16
5

9. Copyright © 2016, 2012 Pearson Education, Inc. 1 -9
Example (continued)
Check:
3
4

16
5
1 ?
7
5
12
5

5
5
7
5
7
5
3
4
x 1 
7
5
The solution is
16
5
.
TRUE

10. Copyright © 2016, 2012 Pearson Education, Inc. 1 -10
Example - Special Case
Solve:
Solution:
24x  7  17  24x
24x  7  17  24
24x  24x  7  24x 17  24x
7 17
Some equations have no solution.
No matter what number we substitute for x, we get a
false sentence.
Thus, the equation has no solution.

11. Copyright © 2016, 2012 Pearson Education, Inc. 1 -11
Example - Special Case
Solve:
Solution:
3
1
3
x  
1
3
x  3
1
3
x  3
1
3
x 
1
3
x 
1
3
x  3
3  3
There are some equations for which any real number
is a solution.
Replacing x with any real number gives a true sentence.
Thus any real number is a solution. The equation has
infinitely many solutions. The solution set is the set of
real numbers, {x | x is a real number}, or (–∞, ∞).

12. Copyright © 2016, 2012 Pearson Education, Inc. 1 -12
Applications Using Linear Models
Mathematical techniques are used to answer questions
arising from real-world situations.
Linear equations and functions model many of these
situations.

13. Copyright © 2016, 2012 Pearson Education, Inc. 1 -13
Five Steps for Problem Solving
1. Familiarize yourself with the problem situation.
Make a drawing Find further information
Assign variables Organize into a chart or table
Write a list Guess or estimate the answer
2. Translate to mathematical language or symbolism.
3. Carry out some type of mathematical manipulation.
4. Check to see whether your possible solution actually
fits the problem situation.
5. State the answer clearly using a complete sentence.

14. Copyright © 2016, 2012 Pearson Education, Inc. 1 -14
The Motion Formula
The distance d traveled by an object moving at
rate r in time t is given by
d = rt.

15. Copyright © 2016, 2012 Pearson Education, Inc. 1 -15
Example
America West Airlines’ fleet includes Boeing 737-
200’s, each with a cruising speed of 500 mph, and
Bombardier deHavilland Dash 8-200’s, each with a
cruising speed of 302 mph (Source: America West
Airlines). Suppose that a Dash 8-200 takes off and
travels at its cruising speed. One hour later, a 737-200
takes off and follows the same route, traveling at its
cruising speed. How long will it take the 737-200 to
overtake the Dash 8-200?

16. Copyright © 2016, 2012 Pearson Education, Inc. 1 -16
Example (continued)
1. Familiarize. Make a drawing showing both the
known and unknown information. Let t = the time, in
hours, that the 737-200 travels before it overtakes the
Dash 8-200. Therefore, the Dash 8-200 will travel t +
1 hours before being overtaken. The planes will travel
the same distance, d.

17. Copyright © 2016, 2012 Pearson Education, Inc. 1 -17
Example (continued)
We can organize the information in a table as follows.
2. Translate. Using the formula d = rt , we get two
expressions for d:
d = 500t and d = 302(t + 1).
Since the distance are the same, the equation is:
500t = 302(t + 1)
Distance Rate Time
737-200 d 500 t
Dash 8-200 d 302 t + 1
d = r • t

18. Copyright © 2016, 2012 Pearson Education, Inc. 1 -18
Example (continued)
500t = 302(t + 1)
500t = 302t + 302
198t = 302
t ≈ 1.53
4. Check. If the 737-200 travels about 1.53 hours, it
travels about 500(1.53) ≈ 765 mi; and the Dash 8-200
travels about 1.53 + 1, or 2.53 hours and travels about
302(2.53) ≈ 764.06 mi, the answer checks.
(Remember that we rounded the value of t).
5. State. About 1.53 hours after the 737-200 has taken
off, it will overtake the Dash 8-200.
3. Carry out. We solve the equation.

19. Copyright © 2016, 2012 Pearson Education, Inc. 1 -19
Simple-Interest Formula
I = Prt
I = the simple interest ($)
P = the principal ($)
r = the interest rate (%)
t = time (years)

20. Copyright © 2016, 2012 Pearson Education, Inc. 1 -20
Example
Jared’s two student loans total $12,000. One loan is at
5% simple interest and the other is at 8%. After 1 year,
Jared owes $750 in interest. What is the amount of
each loan?

21. Copyright © 2016, 2012 Pearson Education, Inc. 1 -21
Example (continued)
Solution:
1. Familiarize. We let x = the amount borrowed at 5%
interest. Then the remainder is $12,000 – x,
borrowed at 8% interest.
Amount
Borrowed
Interest
Rate
Time Amount of Interest
5% loan x 0.05 1 0.05x
8% loan 12,000 – x 0.08 1 0.08(12,000 – x)
Total 12,000 750

22. Copyright © 2016, 2012 Pearson Education, Inc. 1 -22
Example (continued)
2. Translate. The total amount of interest on the two
loans is $750. Thus we write the following equation.
0.05x + 0.08(12,000  x) = 750
3. Carry out. We solve the equation.
0.05x + 0.08(12,000  x) = 750
0.05x + 960  0.08x = 750
 0.03x + 960 = 750
0.03x = 210
x = 7000
If x = 7000, then 12,000  7000 = 5000.

23. Copyright © 2016, 2012 Pearson Education, Inc. 1 -23
Example (continued)
4. Check. The interest on $7000 at 5% for 1 yr is
$7000(0.05)(1), or $350. The interest on $5000 at
8% for 1 yr is $5000(0.08)(1) or $400. Since $350 +
$400 = $750, the answer checks.
5. State. Jared borrowed $7000 at 5% interest and
$5000 at 8% interest.

24. Copyright © 2016, 2012 Pearson Education, Inc. 1 -24
Zeros of Linear Functions
An input c of a function f is called a zero of the
function, if the output for the function is 0 when the
input is c. That is, c is a zero of f if f (c) = 0.
A linear function f (x) = mx + b, with m  0, has
exactly one zero.

25. Copyright © 2016, 2012 Pearson Education, Inc. 1 -25
Example
Find the zero of f(x) = 5x  9.
Algebraic Solution:
5x  9 = 0
5x = 9
x = 1.8
Visualizing the Solution:
The intercept of the graph is
(9/5, 0) or (1.8, 0).